# Using binomial theorem, expand each of the following:

Question:

Using binomial theorem, expand each of the following:

$\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$

Solution:

To find: Expansion of $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

We have, $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$

Let $\left(1+\frac{x}{2}\right)=a$ and $\left(-\frac{2}{x}\right)=b \ldots$ (i)

Now the equation becomes $(a+b)^{4}$

$\Rightarrow\left[{ }^{4} C_{0}(a)^{4-0}\right]+\left[{ }^{4} C_{1}(a)^{4-1}(b)^{1}\right]+\left[{ }^{4} C_{2}(a)^{4-2}(b)^{2}\right]+\left[{ }^{4} C_{3}(a)^{4-3}(b)^{3}\right]+$

$\left[{ }^{4} C_{4}(b)^{4}\right]$

$\Rightarrow\left[{ }^{4} C_{0}(a)^{4}\right]+\left[{ }^{4} C_{1}(a)^{3}(b)^{1}\right]+\left[{ }^{4} C_{2}(a)^{2}(b)^{2}\right]+\left[{ }^{4} C_{3}(a)(b)^{3}\right]+\left[{ }^{4} C_{4}(b)^{4}\right]$

(Substituting value of b from eqn. i)

$\Rightarrow\left[\frac{4 !}{0 !(4-0) !}(a)^{4}\right]+\left[\frac{4 !}{1 !(4-1) !}(a)^{3}\left(-\frac{2}{x}\right)^{1}\right]+\left[\frac{4 !}{2 !(4-2) !}(a)^{2}\left(-\frac{2}{x}\right)^{2}\right]+$

$\left[\frac{4 !}{3 !(4-3) !}(a)^{1}\left(-\frac{2}{x}\right)^{3}\right]+\left[\frac{4 !}{4 !(4-4) !}\left(-\frac{2}{x}\right)^{4}\right]$

(Substituting value of a from eqn. i)

$\left.\Rightarrow\left[1\left(1+\frac{x}{2}\right)\right]^{4}\right]-\left[4\left(1+\frac{x}{2}\right)^{3}\left(\frac{2}{x}\right)\right]+\left[6\left(1+\frac{x}{2}\right)^{2}\left(\frac{4}{x^{2}}\right)\right]$

$-\left[4\left(1+\frac{x}{2}\right)^{1}\left(\frac{8}{x^{3}}\right)\right]+\left[1\left(\frac{16}{x^{4}}\right)\right]$ .....(ii)

We need the value of $a^{4}, a^{3}$ and $a^{2}$, where a $=\left(1+\frac{x}{2}\right)$

For $\left(1+\frac{x}{2}\right)^{4}$ Applying Binomial theorem

$\left(1+\frac{x}{2}\right)^{4}=$

$\left[{ }^{4} C_{0}(1)^{4-0}\right]+\left[{ }^{4} C_{1}(1)^{4}-1\left(\frac{x}{2}\right)^{1}\right]+\left[{ }^{4} C_{2}(1)^{4}-2\left(\frac{x}{2}\right)^{2}\right]+\left[{ }^{4} C_{3}(1)^{4}-\right.$

$\left.3\left(\frac{x}{2}\right)^{3}\right]+\left[{ }^{4} C_{4}\left(\frac{x}{2}\right)^{4}\right]$

$\Rightarrow\left[\frac{4 !}{0 !(4-0) !}(1)^{4}\right]+\left[\frac{4 !}{1 !(4-1) !}(1)^{3}\left(\frac{x}{2}\right)^{1}\right]+\left[\frac{4 !}{2 !(4-2) !}(1)^{2}\left(\frac{x}{2}\right)^{2}\right]$

$+\left[\frac{4 !}{3 !(4-3) !}(1)\left(\frac{x}{2}\right)^{3}\right]+\left[\frac{4 !}{4 !(4-4) !}\left(\frac{x}{2}\right)^{4}\right]$

$\Rightarrow[1]+\left[4(1)\left(\frac{x}{2}\right)\right]+\left[6(1)\left(\frac{x^{2}}{4}\right)\right]+\left[4(1)\left(\frac{x^{3}}{8}\right)\right]+\left[1\left(\frac{x^{4}}{16}\right)\right]$

$\Rightarrow 1+2 x+\frac{3}{2} x^{2}+\frac{x^{3}}{2}+\frac{x^{4}}{16}$

On rearranging the above eqn.

$\Rightarrow \frac{1}{16} x^{4}+\frac{1}{2} x^{3}+\frac{3}{2} x^{2}+2 x+1 \ldots$(iii)

We have, $\left(1+\frac{x}{2}\right)^{4}=\frac{1}{16} x^{4}+\frac{1}{2} x^{3}+\frac{3}{2} x^{2}+2 x+1$

For, $(a+b)^{3}$, we have formula $a^{3}+b^{3}+3 a^{2} b+3 a b^{2}$

For, $\left(1+\frac{x}{2}\right)^{3}$ substituting $a=1$ and $b=\frac{x}{2}$ in the above formula

$\Rightarrow 1^{3}+\left(\frac{x}{2}\right)^{3}+3(1)^{2}\left(\frac{x}{2}\right)+3(1)\left(\frac{x}{2}\right)^{2}$

$\Rightarrow 1+\left(\frac{x^{3}}{8}\right)+\left(\frac{3 x}{2}\right)+\left(\frac{3 x^{2}}{4}\right)$

$\Rightarrow\left(\frac{x^{3}}{8}\right)+\left(\frac{3 x^{2}}{4}\right)+\left(\frac{3 x}{2}\right)+1 \ldots$ (iv)

For, $(a+b)^{2}$, we have formula $a^{2}+2 a b+b^{2}$

For, $\left(1+\frac{x}{2}\right)^{2}$, substituting $a=1$ and $b=\frac{x}{2}$ in the above formula

$\Rightarrow(1)^{2}+2(1)\left(\frac{x}{2}\right)+\left(\frac{x}{2}\right)^{2}$

$\Rightarrow 1+x+\left(\frac{x^{2}}{4}\right)$

$\Rightarrow \frac{x^{2}}{4}+x+1 \ldots(v)$

Putting the value obtained from eqn. (iii),(iv) and (v) in eqn. (ii)

$\Rightarrow\left[1\left(\frac{1}{16} x^{4}+\frac{1}{2} x^{3}+\frac{3}{2} x^{2}+2 x+1\right)\right]-\left[4\left(\frac{x^{3}}{8}+\frac{3 x^{2}}{4}+\frac{3 x}{2}+1\right)\left(\frac{2}{x}\right)\right]$

$\left[6\left(\frac{x^{2}}{4}+x+1\right)\left(\frac{4}{x^{2}}\right)\right]-\left[4\left(1+\frac{x}{2}\right)\left(\frac{8}{x^{3}}\right)\right]+\left[1\left(\frac{16}{x^{4}}\right)\right]$

$\Rightarrow \frac{1}{16} x^{4}+\frac{1}{2} x^{3}+\frac{3}{2} x^{2}+2 x+1-x^{2}-6 x-12-\frac{8}{x}+6+\frac{24}{x}+\frac{24}{x^{2}}$

$-\frac{32}{x^{3}}-\frac{16}{x^{2}}+\frac{16}{x^{4}}$

On rearranging

Ans) $\frac{1}{16} x^{4}+\frac{1}{2} x^{3}+\frac{1}{2} x^{2}-4 x-5+\frac{16}{x}+\frac{8}{x^{2}}-\frac{32}{x^{3}}+\frac{16}{x^{4}}$